What does the "r" in the formula F = - GMm/r^2 mean?

[Miraj Kayastha]

What does the "r" in the formula F = - GMm/r^2 mean?
Does it mean distance between the two bodies or the radial displacement ? Is the "r" a vector or a scalar?

 

[A.T.]

Is the "r" a vector or a scalar?

How do you square a vector? By scalar product with itself? Then it doesn't matter.

 

[Miraj Kayastha]

"dr" is used when potential energy is derived from integration. Here is dr a small displacement in the direction of r?
To calculate the work done by gravity using integration when a body if falling is dr positive or negative?

 

[olivermsun]

$dr$ is infinitesimally small. This allows the value of $F(r)$ to change from one point to the next along $r.$

You have to be a little careful about interpreting the sign of the formula you show above:

If for example, you define $r$ to be positive pointing from the big $M$ to the little $m$, then $F(r)$ in your formula is the force exerted on little $m$ in the $-r$ direction (toward big $M$), and therefore $F(r)$ has negative sign.

Following the same convention, "falling" means moving toward $M$, so $dr$ is also negative. Hence $\int W = \int F dr > 0$, i.e., gravity is exerting force in the same direction as the travel, hence it is doing positive work.

 

[HallsofIvy]

What does the "r" in the formula F = - GMm/r^2 mean?
Does it mean distance between the two bodies or the radial displacement ? Is the "r" a vector or a scalar?

'r' is the distance between the two bodies at a given instant. Since this is a "static" rather than "dynamic" equation, there is NO motion and so no "radial displacement".

r, as used here, is a scalar, the distance between the two objects. If \vec{r} is the vector from one body to the other, then r^2= \vec{r}\cdot\vec{r} or, equivalently, r^2= ||\vec{r}||^2.

Since the direction of the gravitational force on a body is toward the other, we can write this as a vector equation by multiplying by the unit vector from one body to the other, \frac{\vec{r}}{r}:
\vec{F}= -\frac{GMm}{r^2}\frac{\vec{r}}{r}= -\frac{GMm\vec{r}}{r^3}

 

[Miraj Kayastha]

$dr$ is infinitesimally small. This allows the value of $F(r)$ to change from one point to the next along $r.$

You have to be a little careful about interpreting the sign of the formula you show above:

If for example, you define $r$ to be positive pointing from the big $M$ to the little $m$, then $F(r)$ in your formula is the force exerted on little $m$ in the $-r$ direction (toward big $M$), and therefore $F(r)$ has negative sign.

Following the same convention, "falling" means moving toward $M$, so $dr$ is also negative. Hence $\int W = \int F dr > 0$, i.e., gravity is exerting force in the same direction as the travel, hence it is doing positive work.

If dr is negative in case of a falling object, the integration gives a negative result since the lower limit will be greater than the upper limit. So, what's wrong?

 

[nrqed]

If dr is negative in case of a falling object, the integration gives a negative result since the lower limit will be greater than the upper limit. So, what's wrong?

I think it can cause confusion to talk about a negative "dr" since there can be confusion with \vec{dr} (which cannot be "negative") and with |\vec{dr}| (which cannot be negative either!).

Here is another way to present this: the work is

W = \int \vec{F} \cdot \vec{dr}

Now, as an object is falling straight down, the force and the displacement vectors are in the same direction, therefore the dot product is positive and simply equal to |\vec{F} | \, |\vec{dr}|

Now, after integrating, you get - G m M /r (the sign comes from the integration).

CORRECTION: I forgot to mention that since the value of r is decreasing position is decreasing, there is an additional minus sign that we must incorporate. The rest of my post is unchanged.

Evaluating this between the initial and final positions gives a positive answer because 1/r_f is larger than 1/r_i.

 

[olivermsun]

I think it can cause confusion to talk about a negative "dr" since there can be confusion with \vec{dr} (which cannot be "negative") and with |\vec{dr}| (which cannot be negative either!).

You're right, talking about negative dr is confusing. The way I think about it is that dr is negative because the integration is from high to low. But…yeah.

I agree that expressing the integral using the dot product is best. I was just trying to avoid explaining the dot product.

 

[Miraj Kayastha]

I am still not clear about the integration because the work done by a variable force is ∫F.dx where dx is a very small DISPLACEMENT.

Whereas while calculating the work done by the gravitational force we do ∫F.dr where dr is a radial displacement that is always outwards. So is dr equal to dx? I mean dr is not the displacement if a body is falling, right?

 

[nrqed]

I am still not clear about the integration because the work done by a variable force is ∫F.dx where dx is a very small DISPLACEMENT.

Whereas while calculating the work done by the gravitational force we do ∫F.dr where dr is a radial displacement that is always outwards. So is dr equal to dx? I mean dr is not the displacement if a body is falling, right?

I am sorry, I forgot to mention an important point in my previous post (which I corrected). Let me be more clear.

We have

W = \int_{r_i}^{r_f} \vec{F} \cdot \vec{dr} = + \int_{r_i}^{r_f} | \vec{F} | ~ | \vec{dr} | <br />

Now, | \vec{F} | = G m M /r^2

We must be careful with | \vec{dr} |. We need to relate this to dr, which is the change of value in the coordinate r. As Olivermsum correctly mentioned, since the value of r is decreasing, the change of coordinate in r is negative so | \vec{dr} | = - dr.

So putting everything together, we get

+ \int_{r_i}^{r_f} | \vec{F} | ~ | \vec{dr} | = - \int_{r_i}^{r_f} \frac{G m M}{r^2} dr

Doing this integral, we get

<br /> + \frac{G mM}{r_f} - \frac{GmM}{r_i}
And this is positive because r_i &gt; r_ f

 

[Khashishi]

r is the distance between the centers of the two objects which are assumed to be either point particles or spherical particles. For weirdly shaped objects, you can take r to be the distance between the centers of mass, but the equation "F = - GMm/r^2" is only approximately true for this case. The approximation is good if the distance r is large compared to the sizes of the objects and/or the objects are almost spherically symmetric in mass distribution.

 

[Miraj Kayastha]

I am sorry, I forgot to mention an important point in my previous post (which I corrected). Let me be more clear.

We have

W = \int_{r_i}^{r_f} \vec{F} \cdot \vec{dr} = + \int_{r_i}^{r_f} | \vec{F} | ~ | \vec{dr} | <br />

Now, | \vec{F} | = G m M /r^2

We must be careful with | \vec{dr} |. We need to relate this to dr, which is the change of value in the coordinate r. As Olivermsum correctly mentioned, since the value of r is decreasing, the change of coordinate in r is negative so | \vec{dr} | = - dr.

So putting everything together, we get

+ \int_{r_i}^{r_f} | \vec{F} | ~ | \vec{dr} | = - \int_{r_i}^{r_f} \frac{G m M}{r^2} dr

Doing this integral, we get

<br /> + \frac{G mM}{r_f} - \frac{GmM}{r_i}
And this is positive because r_i &gt; r_ f

Why did you put the magnitude of force, shouldn't it contain minus sign as well? Because F is in opposite direction to r?

I read a post online that said if we state -dr we are already assuming the motion of the body is downwards so the limits should be switched and if we state dr then only we can keep the initial value on the lower limit and final value at the upper limit because if we keep dr positive then the limits determine the direction of motion rather than the dr?
Is this true?

The article I read : http://www.wired.com/2012/03/deriving-the-gravitational-potential-energy/

Also, is the derivation of GPE different for falling and rising objects different because in falling dr is negative whereas in rising dr is positive?